Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-8x-2y &= -1 \\ 9x+y &= 6\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $9x = -y+6$ Divide both sides by $9$ to isolate $x$ $x = {-\dfrac{1}{9}y + \dfrac{2}{3}}$ Substitute this expression for $x$ in the first equation. $-8({-\dfrac{1}{9}y + \dfrac{2}{3}}) - 2y = -1$ $\dfrac{8}{9}y - \dfrac{16}{3} - 2y = -1$ Simplify by combining terms, then solve for $y$ $-\dfrac{10}{9}y - \dfrac{16}{3} = -1$ $-\dfrac{10}{9}y = \dfrac{13}{3}$ $y = -\dfrac{39}{10}$ Substitute $-\dfrac{39}{10}$ for $y$ in the top equation. $-8x-2( -\dfrac{39}{10}) = -1$ $-8x+\dfrac{39}{5} = -1$ $-8x = -\dfrac{44}{5}$ $x = \dfrac{11}{10}$ The solution is $\enspace x = \dfrac{11}{10}, \enspace y = -\dfrac{39}{10}$.